Carrying Capacity - Biology

1. Fundamental Concepts

Definition: Carrying capacity is the maximum population size of a species that an environment can sustain indefinitely, given the food, habitat, water, and other necessities available in the environment.
Factors Influencing Carrying Capacity: Food availability, water supply, shelter, space, and predation are key factors that determine the carrying capacity of an ecosystem.
Population Dynamics: Population growth tends to slow down as it approaches the carrying capacity due to resource limitations.

2. Key Concepts

Logistic Growth Model: $$\frac{dN}{dt} = rN \left(1 - \frac{N}{K}\right)$$

Where $N$ is the population size, $r$ is the intrinsic growth rate, and $K$ is the carrying capacity.

Environmental Resistance: The concept explains how environmental factors limit population growth.

Application: Used in ecology to predict population sizes and manage resources.

3. Examples

Example 1 (Basic)

Problem: A population of rabbits has an intrinsic growth rate ($r$) of 0.5 per month and a carrying capacity ($K$) of 200. If the initial population ($N_0$) is 50, what will the population be after one month?

Step-by-Step Solution:

Substitute the values into the logistic growth equation: $$N(t) = \frac{K N_0 e^{rt}}{K + N_0 (e^{rt} - 1)}$$
Calculate for $t = 1$: $$N(1) = \frac{200 \cdot 50 \cdot e^{0.5 \cdot 1}}{200 + 50 (e^{0.5 \cdot 1} - 1)}$$
Simplify: $$N(1) = \frac{10000 \cdot e^{0.5}}{200 + 50 (e^{0.5} - 1)}$$
Compute: $$N(1) \approx 76.92$$

Validation: The calculated population size is within the expected range based on the given parameters.

Example 2 (Intermediate)

Problem: Given a population with an intrinsic growth rate ($r$) of 0.3 per year and a carrying capacity ($K$) of 500. If the current population ($N$) is 250, estimate the time ($t$) it takes for the population to reach 400.

Step-by-Step Solution:

Rearrange the logistic growth equation to solve for $t$: $$t = \frac{1}{r} \ln \left(\frac{K N_0}{K - N_0}\right)$$
Substitute the values: $$t = \frac{1}{0.3} \ln \left(\frac{500 \cdot 250}{500 - 250}\right)$$
Simplify: $$t = \frac{1}{0.3} \ln \left(\frac{125000}{250}\right)$$
Compute: $$t \approx 4.62 \text{ years}$$

Validation: The estimated time aligns with the expected growth pattern given the parameters.

4. Problem-Solving Techniques

Graphical Analysis: Plot population size against time to visualize growth patterns and identify trends.
Dimensional Consistency: Ensure all units are consistent when substituting values into equations.
Estimation Checks: Use rough estimates to check if the final answer is reasonable.